On transforming pair of bivariate polynomials to pair of univariate polynomials by applying polynomial map We know that a polynomial map $f(x,y), g(x,y)$ is polynomial automorphism if there exists polynomials $p(x,y)$ and $q(x,y)$ such that $f(p,q)$=x and $g(p,q)=y$. Jacobian conjecture tries to characterize polynomial automorphisms.
My question is, for any two algebraically independent bivariate polynomials $f$ and $g$, can we always find polynomials or rational functions $p(x,y)$ and $q(x,y)$ such that $f(p,q)$ is a univariate polynomial in $x$ and $g(p,q)$ is a univariate polynomial in $y$. If the answer is no, i want a counterexample and characterization of polynomials maps for which there exist such $p$ and $q$. Even if we cannot convert to univariate for some $f$ and $g$, can we always convert to univariate by applying polynomial map on two polynomials in the ring generated by two algebraically independent polynomials $f(x,y)$ and $g(x,y)$
 A: Let $k$ be an algebraically closed field, e.g., $\mathbb{C}$.
Given algebraically independent polynomials $f(u,v)$ and $g(u,v)$ in $k[u,v]$, it is typically impossible to find algebraically independent rational functions $p(x,y)$ and $q(x,y)$ in $k(x,y)$ such that $f(p(x,y),g(x,y))$ is univariate in $x$ and $g(p(x,y),q(x,y))$ is univariant in $y$.  Of course if you drop the condition that $p$ and $q$ should be algebraically independent, you can trivially arrange this by choosing both $p$ and $q$ to be constant polynomials, but I imagine you want to exclude this case.
For typical $f$ and $g$, for a typical element $c$ in $k$, the solution set of $f(u,v) = c$ is an affine, irreducible curve $Z$ of some geometric genus $g$ (by "geometric genus", I mean the genus of any smooth, projective model of this affine curve).  If $g$ is positive, there is no nonconstant polynomial map from any curve whose geometric genus is $0$ to $Z$.  If $f(p,q)$ is univariate, then the solution set of $f(p(x,y),q(x,y)) = c$ is a union of copies of the $y$-axis, every component of which has geometric genus $0$.  The restriction of the polynomial map $(p(x,y),q(x,y))$ to these copies of the $y$-axis give polynomial maps from these curves of geometric genus $0$ to $Z$.  This map must be constant.  Since this holds for general $c$, the polynomial map $(p(x,y),q(x,y))$ is not dominant (or else it would be generically finite as well).  Therefore $p(x,y)$ and $q(x,y)$ are algebraically dependent.
